Question #1841
A small domestic elevator has a mass of 454 kg and can ascend at a rate of 0.180 m/s. What is the average power that must be supplied for the elevator to move at this rate? answer in Watts.
Expert's answer
The power is equal to W = A/t
The work done by the elevator is equal to changing fo the potential energy:& mgΔH
thus W = mgΔH/t = mg v = 454[kg]*10[m/s2] * 0.18 [m/s] = 817.2 W.
The work done by the elevator is equal to changing fo the potential energy:& mgΔH
thus W = mgΔH/t = mg v = 454[kg]*10[m/s2] * 0.18 [m/s] = 817.2 W.
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#339914 on Dec 2023
#339914 on Dec 2023